http://acm.hdu.edu.cn/showproblem.PHP?pid=1002
A + B Problem II
Problem Description
I have a very simple problem for you. Given two integers A and B,your job is to calculate the Sum of A + B.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow,each line consists of two positive integers,A and B. Notice that the integers are very large,that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
Output
For each test case,you should output two lines. The first line is "Case #:",# means the number of the test case. The second line is the an equation "A + B = Sum",Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
2 1 2 112233445566778899 998877665544332211
Sample Output
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
第一次算大数相加,在网上看了思想,自己敲代码,由于粗心数组初始化出了点问题,提交了n次都没过,还好细心的队友发现啦!
#include<iostream> #include<cstdio> #include<string.h> #define N 1005 int main() { int i,j,k,m,l1,l2,t; char s1[N],s2[N]; int a[N],b[N]; scanf("%d",&t); for(m=1;m<=t;m++) { memset(s1,sizeof(s1)); memset(s2,sizeof(s2)); memset(a,sizeof(a)); memset(b,sizeof(b)); scanf("%s%s",&s1,&s2); l1=strlen(s1); l2=strlen(s2); if(l1>l2) k=l1; else k=l2; for(i=k,j=l1-1;j>=0;i--,j--) a[i]=s1[j]-'0'; for(i=k,j=l2-1;j>=0;i--,j--) b[i]=s2[j]-'0'; for(i=k;i>0;i--) { a[i]+=b[i]; if(a[i]>=10) { a[i]=a[i]%10; a[i-1]++; } } printf("Case %d:\n%s + %s = ",s1,s2); if(a[0]!=0) { for(i=0;i<=k;i++) { printf("%d",a[i]); } } else { for(i=1;i<=k;i++) printf("%d",a[i]); } if(m<=t-1) printf("\n\n"); else printf("\n"); } return 0; }